Cartesian product question

Okay, I have this problem from a maths textbook and its along the same lines as

If S and T are sets

the Cartesian product S x T of S and T is defined


[(s,t) : s ∈ S, t ∈ T}


Prove if S and T are convex sets in ℝ^n and ℝ^m

then ℝ^(n+m) is convex also

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I have the answer but I'm completely confused on how it was derived
 
Jul 2009
193
52
Suppose S and T are convex sets in \(\displaystyle \Re ^n\) and \(\displaystyle \Re ^m\) respectively
\(\displaystyle \Longleftrightarrow\) \(\displaystyle \forall x_1,y_1 \in S\), \(\displaystyle \forall t \in [0,1]\)
\(\displaystyle (1-t)x_1 + ty_1 \in \Re ^n\)
and
\(\displaystyle \forall x_2,y_2 \in T\), \(\displaystyle \forall t \in [0,1]\)
\(\displaystyle (1-t)x_2 + ty_2 \in \Re ^m\)

with \(\displaystyle a(x,y)\) defined as \(\displaystyle (ax,ay)\) for \(\displaystyle a \in \Re\) and \(\displaystyle (x,y)\in S \times T\) and \(\displaystyle (x_1,y_1) + (x_2,y_2) = (x_1+x_2,y_1+y_2)\)...
 
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Okay, I don't fully understand your answer, heres what I have;

λs1 + (1-λ)s2 \(\displaystyle forall\) S \(\displaystyle forall R^n\)

λt1 + (1-λ)t2 \(\displaystyle forall \) T \(\displaystyle forall R^m\)

and now I can't understand why R^n , R^m makes SxT convex

thanks
 
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Jul 2009
193
52
Ok so this is what you've written;
λs1 + (1-λ)s2 \(\displaystyle \forall\) S \(\displaystyle \forall R^n\)

λt1 + (1-λ)t2 \(\displaystyle \forall \) T \(\displaystyle \forall R^m\)



s1 and s2 are elements of set S and t1 and t2 are elements of set T. I wouldn't write the "forall Rn" part since the statements are only true for the convex sets S and T in Rn,Rm. So basically S and T are convex sets in \(\displaystyle \Re^n\) and \(\displaystyle \Re^m\). This means that the following by definition is true;
λs1 + (1-λ)s2 \(\displaystyle \in \Re^n\) \(\displaystyle \forall s1,s2 \in\) S

λt1 + (1-λ)t2 \(\displaystyle \in \Re^m\)\(\displaystyle \forall t1,t2 \in\) T

(If S is a convex set in \(\displaystyle \Re^n\) then λs1 + (1-λ)s2 is in \(\displaystyle \Re^n\)).

We've just applied the definition of a convex set so far. Now we want to show the set S \(\displaystyle \times\) T (which is of the form x1=(s1,t1)) is also a convex set. So we want to show that;

λx1 + (1-λ)x2 \(\displaystyle \in \Re^{n \times m}\)\(\displaystyle \forall x1,x2 \in\) S \(\displaystyle \times \) T

We begin with the equation

λx1 + (1-λ)x2\(\displaystyle \forall x1,x2 \in\) S \(\displaystyle \times \) T

which becomes (x1 = (s1,t1), x2 = (s2,t2) are elements of \(\displaystyle S \times T\))

λ(s1,t1) + (1-λ)(s2,t2) \(\displaystyle \forall s1,s2 \in\) S and \(\displaystyle \forall t1,t2 \in\) T

which means (λ(s1,t1) = (λs1,λt1))

(λs1,λt1) + ((1-λ)s2,(1-λ)t2)\(\displaystyle \forall s1,s2 \in\) S and \(\displaystyle \forall t1,t2 \in\) T

which then becomes ((s1,t1)+(s2,t2) = (s1+s2,t1+t2))

((λs1 + (1-λ)s2),(λt1 + (1-λ)t2))\(\displaystyle \forall s1,s2 \in\) S and \(\displaystyle \forall t1,t2 \in\) T

what can we say about the equations (λs1 + (1-λ)s2) and (λt1 + (1-λ)t2) ?
 
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what can we say about the equations (λs1 + (1-λ)s2) and (λt1 + (1-λ)t2) ?
Well am I correct in saying that ((s1,t1)+(s2,t2) = (s1+s2,t1+t2)) therefore S and T are convex IF they both lie in S x T (on a graph)
(I'm slightly confused)

I really have no idea


Edit: If S and T are two convex sets in R^n and R^m then their intersection S∩T is also convex, - I understand that but I don't know how its proved that its convex in R^n+m
 
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Jul 2009
193
52
what can we say about the equations (λs1 + (1-λ)s2) and (λt1 + (1-λ)t2) ?
By definition (written above) (λs1 + (1-λ)s2) is in \(\displaystyle \Re^n\) and (λt1 + (1-λ)t2) is in \(\displaystyle \Re^m\) so that the element ((λs1 + (1-λ)s2),(λt1 + (1-λ)t2)) is in \(\displaystyle S \times T\). This means by definition that \(\displaystyle S \times T\) is a convex set in \(\displaystyle \Re^{n\times m}\)

What we've done is taken any two elements x1 and x2 of \(\displaystyle S \times T\) and showed that λx1 + (1-λ)x2 also lies in \(\displaystyle S \times T\).

Well am I correct in saying that ((s1,t1)+(s2,t2) = (s1+s2,t1+t2)) therefore S and T are convex IF they both lie in S x T (on a graph)
No the operation (s1,t1)+(s2,t2) = (s1+s2,t1+t2) is just how the elements are added. What we need to show is that for any two elements of S x T the equation λx1 + (1-λ)x2 lies in it as well.
 
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